Help help pls pls help help

Answer:

Step-by-step explanation:

m angle APD.

Hope It's help you.

Step-by-step explanation:

yeah angle m is APD so yeah

Answer:

[tex]a_{n}[/tex] = 5n

Step-by-step explanation:

There is a common difference d between consecutive terms

d = 10 - 5 = 15 - 10 = 20 - 15 = 25 - 20 = 30 - 25 = 5

This indicates the sequence is arithmetic with explicit formula

[tex]a_{n}[/tex] = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 5 and d = 5 , then

[tex]a_{n}[/tex] = 5 + 5(n - 1) = 5 + 5n - 5 = 5n

Step-by-step explanation:

Given that,

The bag of skittles has 3 parts red skittles, 5 parts yellow skittles and 7 parts orange skittle.

Ratio notation is= 3:5:7

Frction notation,

Red skittle[tex]=\dfrac{3}{3+5+7}=\dfrac{1}{5}[/tex]

Yellow skittle [tex]=\dfrac{5}{3+5+7}=\dfrac{1}{3}[/tex]

Orange skittle [tex]=\frac{7}{3+5+7}=\dfrac{7}{15}[/tex]

Hence, this is the required solution.

Answer:

B

Step-by-step explanation:

9 divided by 4, then multiply by 12

Answer:

y = 2.25(12)

Step-by-step explanation:

$9 for 4 pizzas

$9/4 = $2.25

$9 for 4 pizzas is the same ratio as $2.25 per 1 pizza

To deliver x number of pizzas, the charge, y, is

y = 2.25x

For 12 pizzas, the delivery charge is

y = 2.25(12)

D. 7

Hope this helps! :)

______________

Answer:

The cube root of 343 is 7.

Answer:

Step 1: Set up a product of two ( ) where each will hold two terms. Step 2: Find the factors that go in the first positions. Step 3: Find the factors that go in the last positions. We need two numbers whose product is 9 and sum is 2.p explanation:

Answer:

A❤️✌️❤️✌️❤️Aaaaaa

Answer:

B. Leg-Acute (LA)

E. Angle-Angle-Side (AAS)

Step-by-step explanation:

Congruence is the relationship between two or more shapes with respect to their common properties.

Comparing the properties of triangles LMN and OPQ, it would be observed that two angles are similar and one side.

So that the congruence theorems or postulates required are:

B. Leg-Acute (LA)

E. Angle-Angle-Side (AAS)

Answer:

Step-by-step explanation:

PERIMETER OF ORIGINAL RECTANGLE

perimeter = 2 × ( length + width)

= 2 × ( 4.5 + 4.3)

= 2 × 8.8

= 17. 6 mm

SIDE LENGTH OF ENLARGED RECTANGLE

Enlargement factor = 24.3 / 4.5

= 5.4

Side length of enlarged rectangle

= original length × Enlargement factor

= 4.3 × 5.4

= 23. 22 mm

PERIMTER OF ENLARGED RECTANGLE

= 2 × ( enlarged ength + enlarged breadth)

= 2 × (23. 22 + 24. 3 )

= 95. 04 mm

Answer:

[tex]{ \tt{ \sqrt[3]{24 {n}^{2} } \times \sqrt[3]{36 {n}^{2} } }} \\ = { \tt{( {n}^{ \frac{2}{3} })( \sqrt[3]{864}) }}[/tex]

Answer:

x(x-x[tex]y^{2}[/tex]+y)

Step-by-step explanation:

Answer:

8 people watched the show

Step-by-step explanation:

50 - 42 = 8

Answer:

2x + 13

13

Step-by-step explanation:

Albert and Maxim own

x + (x + 13) = 2x + 13 model airplanes together

Albert owns

(x + 13) - x = 13 less model airplanes than Maxim.

Answer:

They own 2x+13 model airplanes together

Albert owns 13 less model airplanes than Maxim

Answer:

angle 1 = 105º

angle 2  = 75º

Step-by-step explanation:

angle 2  = 75º

corresponding angle to the 75º angle

angle 1 = 105º

it's supplementary to angle 2

180 - 75 = 105

Answer:

[tex]1\frac{6}{9}[/tex]

Step-by-step explanation:

[tex]\frac{8}{9}+\frac{7}{9}[/tex]

[tex]\frac{8+7}{9}=\frac{15}{9}=1\frac{6}{9}[/tex]

Answer:

x=7/13, y=19/13, z=74/13

Step-by-step explanation:

Add or subtract multiples of equations to eliminate one variable. I picked z to eliminate.

Subtract the 1st equation from the 2nd equation.

2x + 2Y = 4 or x + y = 2 or y = -x + 2, then

Multiple the 2nd equation by 2 and subtract the 3rd equation.

9x - 4y = 5 then sub (-x + 2) for y from previous step.

9x - 4(-x +2) = 5

13x -2 = 5

13x = 7

x = 7/13

plug that into either of the step 1 equations to gey y.

y = -x + 2

y = -7/13 + 2

y = 19/13

then plug the found x and y into any one of the original equations to get z.

-x + 2y - 2z = -9

-7/13 + 2(19/13) - 2z = -9

31/13 - 2z = -9

-2z = -9 - 31/13

-2z = -117/13 - 31/13 = -148/13

z = 74/13

Answer:

Multiply the second equation by 3. The solution is x = 12, y = 10.

Step-by-step explanation:

The given equations are :

x+3y= 42 ....(1)

2x - y = 14 ...(2)

For elimination method,

Multiply the second equation by 3.

6x-3y=42 .....(3)

Adding equations (1) and (3).

x+3y+6x-3y=42+42

7x = 84

x = 12

Put the value of x in equation (1).

12+3y= 42

3y = 30

y = 10

So, x = 12 and y = 10. Hence, the correct option is (d).

Answer:

x=5

y=4

Step-by-step explanation:

Step 1:

Let us solve for x using the [tex]4x-7=13[/tex] equation by isolating terms containing the variable:

[tex]4x-7=13\\4x=20\\\fbox{x=5}[/tex]

Great! x is 5... let's find y.

Step 2:

Since we know the value of x, we can use what we know to find y using the other equation:

[tex]5x-2y=17\\[/tex]

Given that x=5, we can replace x with 5.

[tex]5*5-2y=17\\25-2y=17\\2y=8\\\\\fbox{y=4}[/tex]

I hope this helps! Let me know if you have any questions :)

Answer:

63

Step-by-step explanation:

A=bh=7·9=63

base times perpendicular height

Answer:

63

Step-by-step explanation:

7x9=63

you copied my answer on a different question.

and literally copied and pasted it and it's so obvious and i saw u change it

Answer:

Answer is 12

Step-by-step explanation:

ok 3x2=6 1ñ 2 12

ok 3

Answer:

195 sq cm

Step-by-step explanation:

15 x 15 = 225

----------------------

3(8) + 2(3) = 30

----------------------

225 - 30 = 195 sq cm

Answer:

your answer is 1275

Step-by-step explanation:

ccifufbfjrorwfeejeidiffgfkxxocyegdo

Answer:

5

Step-by-step explanation:

Greater than 0

Less than 100

Is not 4, but is greater than four

Is less than 5.5

Must be a whole number

5 is the only whole number between 4 and 5.5

Answer:

first question answer is step 2 and second step is answer step 3, last one is step is answer step 1

Answer:

[tex]{ \tt{a < b→true}} \\ { \bf{reason :because \: - 1 \: is \: less \: than \: 4.5 }}\\ \\ { \tt{ |a| > b →false}} \\ { \bf{reason :1 \: is \: not \: greater \: than \: 4.5 }}\\ \\ { \tt{a < |b| →true}}[/tex]

Answer:

true

false

true

Step-by-step explanation:

Step-by-step explanation: in the question it is given that, Fred is five times as old as Grace, it means that the present age of Fred is 5 times the age of Grace. in the solution we have assumed Grace age to be X years. if Grace age is X years then the age of Fred is five times that of Greece age that is 5x years. now it is given in the question that in 8 years of time Fred's age will be three times as old as Grace. as it is given in the question as 8 years then it means that we have to add 8 years to both the present age of grace as well as Fred's age. now if we'll add 8 years in both grace age and frads age then it will be X + 8 years and 5 x + 8 years respectively. now it is said that in 8 years of time Fred's age will be three times of grace age. now to make it equal we have to put that three times in Grace age, so that three times of grace age will be equal to Fred's age. then you simply need to solve the equation As given in the solution then we will get X equal to 8. we have got Grace present age to be 8 years and Fred's present age to be 40 years uhmmmmmmmmmmmmmm i hope this hepls you do your work :D

Answer:

4 : 1

Step-by-step explanation:

To convert 2/20 : 1/40 to unit rate, we simplify the left side. So.

2/20 : 1/40 = 1/10 : 1/40

We now multiply both sides by 40.

So, 1/10 : 1/40 = 40 × 1/10 : 40 × 1/40 = 4 : 1

So, our unit rate is 4 : 1

Answer:

y=5, y=[tex]\frac{38}{11}[/tex]

Step-by-step explanation:

Hi there!

We are given the equation

[tex]\frac{y+2}{y-3}[/tex]+[tex]\frac{y-1}{y-4}[/tex]=[tex]\frac{15}{2}[/tex] and we need to solve for y

first, we need to find the domain, which is which is the set of values that y CANNOT be, as the denominator of the fractions cannot be 0

which means that y-3≠0, or y≠3, and y-4≠0, or y≠4

[tex]\frac{y+2}{y-3}[/tex] and [tex]\frac{y-1}{y-4}[/tex] are algebraic fractions, meaning that they are fractions (notice the fraction bar), but BOTH the numerator and denominator have algebraic expressions

Nonetheless, they are still fractions, and we need to add them.

To add fractions, we need to find a common denominator

One of the easiest ways to find a common denominator is to multiply the denominators of the fractions together

Let's do that here;

on [tex]\frac{y+2}{y-3}[/tex], multiply the numerator and denominator by y-4

[tex]\frac{(y+2)(y-4)}{(y-3)(y-4)}[/tex]; simplify by multiplying the binomials together using FOIL to get:

[tex]\frac{y^{2}-2y-8}{y^{2}-7y+12}[/tex]

Now on [tex]\frac{y-1}{y-4}[/tex], multiply the numerator and denominator by y-3

[tex]\frac{(y-1)(y-3)}{(y-4)(y-3)}[/tex]; simplify by multiplying the binomials together using FOIL to get:

[tex]\frac{y^{2}-4y+3}{y^{2}-7y+12}[/tex]

now add [tex]\frac{y^{2}-2y-8}{y^{2}-7y+12}[/tex] and [tex]\frac{y^{2}-4y+3}{y^{2}-7y+12}[/tex] together

Remember: since they have the same denominator, we add the numerators together

[tex]\frac{y^{2}-2y-8+y^{2}-4y+3}{y^{2}-7y+12}[/tex]

simplify by combining like terms

the result is:

[tex]\frac{2y^{2}-6y-5}{y^{2}-7y+12}[/tex]

remember, that's set equal to [tex]\frac{15}{2}[/tex]

here is our equation now:

[tex]\frac{2y^{2}-6y-5}{y^{2}-7y+12}[/tex]=[tex]\frac{15}{2}[/tex]

it is a proportion, so you may cross multiply

2(2y²-6y-5)=15(y²-7y+12)

do the distributive property

4y²-12y-10=15y²-105y+180

subtract 4y² from both sides

-12y-10=11y²-105y+180

add 12 y to both sides

-10=11y²-93y+180

add 10 to both sides

11y²-93y+190=0

now we have a quadratic equation

Let's solve this using the quadratic formula

Recall that the quadratic formula is y=(-b±√(b²-4ac))/2a, where a, b, and c are the coefficients of the numbers in a quadratic equation

in this case,

a=11

b=-93

c=190

substitute into the formula

y=(93±√(8649-4(11*190))/2*11

simplify the part under the radical

y=(93±√289)/22

take the square root of 289

y=(93±17)/22

split into 2 separate equations:

y=[tex]\frac{93+17}{22}[/tex]

y=[tex]\frac{110}{22}[/tex]

y=5

and:

y=[tex]\frac{93-17}{22}[/tex]

y=[tex]\frac{76}{22}[/tex]

y=[tex]\frac{38}{11}[/tex]

Both numbers work in this case (remember: the domain is y≠3, y≠4)

So the answer is:

y=5, y=[tex]\frac{38}{11}[/tex]

Hope this helps! :)

[tex]\underline \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }[/tex]

[tex]\huge\underline{\sf{\red{Problem:}}}[/tex]

[tex]\underline{\sf{\red{Formula\:for\: area\: of \:rectangular\: form:}}}[/tex]

[tex]\quad\quad\quad\quad\boxed{\sf{\red{➢} \: \sf{a=l×b}} }[/tex]

[tex]\underline{\sf{\red{Given:}}}[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{Length\:of\:rectangular\:form = 45m} }[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{Breadth\:of\:rectangular\:form = 15m} }[/tex]

[tex]\underline{\sf{\red{Solution:}}}[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{a=l×b} }[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{a=45m×15m} }[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{a=\boxed{\sf{{675m}^{2}}}} }[/tex]

[tex]\underline{\sf{\red{For\:rectangular\: form\:with\:path:}}}[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{Length\:of\:rectangular\:form = 45m+2=47m} }[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{Breadth\:of\:rectangular\:form = 15m+2=17m} }[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{a=l×b} }[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{a=47m×17m} }[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{a=\boxed{\sf{{799m}^{2}}}} }[/tex]

[tex]\huge\underline{\sf{\red{Area\:of\:path:}}}[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{a={799m}^{2}-{675m}^{2}} }[/tex]

[tex]\quad\quad\quad\quad\boxed{\sf{\red{➢} \: \sf{a={124m}^{2}} }}[/tex]

[tex]\huge\underline{\sf{\red{Answer:}}}[/tex]

[tex]\huge\quad\quad\underline{\boxed{\sf{\red{a={124m}^{2}} }}}[/tex]

[tex]\underline \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }[/tex]

[tex]\sf{\red{︎✰}ShineBrighter\red{︎✰}}[/tex]

[tex]\sf{✍︎ C.Rose\red{❀}}[/tex]

ฅ^•ﻌ•^ฅ